我想为我的ForigenKey列而不是数值显示文本。有很多示例通过比较ID来检索TextMember,但它们不适用于我的案例。我刚开始使用剑道UI所以不要很了解剑道网格。如何在Grid的单元格中显示dataTextField而不是dataValueField?
下面是代码:
$(document).ready(function() {
dataSource1 = new kendo.data.DataSource({
transport: {
read: {
url: "Data/AttendanceCode/GridSelect.php",
dataType: "json",
},
update: {
url: "Data/AttendanceCode/GridUpdate.php",
dataType: "json",
type:"GET"
},
destroy: {
url: "Data/AttendanceCode/GridDelete.php",
dataType: "json",
type:"POST"
},
create: {
url: "Data/AttendanceCode/GridInsert.php",
dataType: "json",
type:"POST"
},
},
schema: {
data: "data",
model: {
id: "AttendenceID",
fields: {
AttendenceID : { editable: false, nullable: true },
TeacherID: { field: "TeacherID", defaultValue: "EIIT0002" },
}
}
},
});
$("#grid").kendoGrid({
dataSource: dataSource1,
pageSize: 10,
pageable: {
refresh: true,
pageSizes: true
},
editable:{ mode : "popup" },
height: 400,
filterable: true,
columnMenu: true,
sortable: true,
reorderable: true,
resizable: true,
toolbar: ["create"],
columns: [
{ field:"AttendenceID", title: "Attendence ID", width:"130px" },
{ field: "TeacherID", title:"Teacher", width: "100px" , editor: TeacherDropDownEditor, template: "#=getTeacherName(TeacherID)#" },
{ command: ["edit", "destroy"], title: "Action", width: "210px" }],
});
});
教师下拉数据源
teacher = new kendo.data.DataSource({
transport: {
read: {
url : "Data/Teacher.php",
dataType: "json" }
},
schema: {
data : "Teacher"
}
});
//教师主编
function TeacherDropDownEditor(container, options) {
$('<input data-bind="value:' + options.field + '"/>')
.appendTo(container)
.kendoDropDownList({
dataTextField: "TeacherName",
dataValueField: "Service_NO",
dataSource: teacher
});
}
我发现并试图获得教师姓名的不同方法
1 -
function getTeacherName(value) {
var text = "";
$.each(teacher, function() {
if (this.Service_NO == value) {
text = this.Name;
return false;
}
});
return text;
}
2 -
function getTeacherName(teacherID) {
for (var idx = 0, length = teacher.length; idx < length; idx++)
{
if (teacher[idx].Service_NO === teacherID)
{t = teacher[idx].Name;}
}
return t;
}
3 -
function getTeacherName(teacherID) {
$.each(teacher, function(key, val) {
if(val.Service_NO == tID){
t = val.Name;
}
});
return t;
}
好像数据源(教师)不具有任何值。 PHP代码运行完美。 如果您对我的代码有什么疑问,请帮助。
谢谢!
代码工作。谢谢OnaBai ... :-) –